#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (iii) textbook solution.

Answer : $y+2 x=3 x^{2} y$

Hint               : using variable separable method and substituting the values

Given :   $x^{2} d y+\left(x y+y^{2}\right) d x=0, \quad y=1, \text { when } x=1$

Solution : $x^{2} d y+\left(x y+y^{2}\right) d x=0$

The differential equation can be  written as

\begin{aligned} &x^{2} d y=-\left(x y+y^{2}\right) d x\\ &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}                                      ......(i)

Let $f(x, y)=\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}}$

Finding  $f(\lambda \; x,\lambda \; y)$

\begin{aligned} f(\lambda x, \lambda y) &=\frac{-\left(\lambda x . \lambda y+\lambda^{2} y^{2}\right)}{\lambda^{2} x^{2}}=\frac{-\lambda^{2}\left(x y+y^{2}\right)}{\lambda^{2} x^{2}} \\ &=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &=\lambda^{0}(f(x, y)) \\ f(x, y)=& \frac{-\left(x y+y^{2}\right)}{x^{2}} \end{aligned}

Therefore , $f(x,y)$ is a homogenous function of degree zero.

Putting $y=vx$

Diff w.r.t.x

$\frac{d y}{d x}=x \frac{d v}{d x}+v$

Putting value of $\frac{d y}{d x} \text { and } y=v x \text { in }(i)$

\begin{aligned} &\frac{d y}{d x}=\frac{-\left(x y+y^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x(v x)+(v x)^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-\left(x^{2} v+x^{2} v^{2}\right)}{x^{2}} \\ &v+x \frac{d v}{d x}=\frac{-x^{2}\left(v+v^{2}\right)}{x^{2}} \end{aligned}

\begin{aligned} &x \frac{d v}{d x}=-v-v^{2}-v=-\left(v^{2}+2 v\right) \\ &\frac{d v}{v^{2}+2 v}=-\frac{d x}{x} \end{aligned}

Integrating both sides we get ,

\begin{aligned} &\int \frac{d v}{v^{2}+2 v}=-\int \frac{d x}{x} \\ \end{aligned}

\begin{aligned} &\int \frac{d v}{\left(v^{2}+2 v+1\right)-1}=-\log x+\log c \\ \end{aligned}

\begin{aligned} &\int \frac{d v}{(v+1)^{2}-1^{2}}=-\log x+\log c \\ \end{aligned}

Using \begin{aligned} & \int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c \\ \end{aligned}

\begin{aligned} &\frac{1}{2} \log \frac{v+1-1}{v+1+1}=-\log x+\log c \end{aligned}

\begin{aligned} &\frac{1}{2} \log \frac{v}{v+2}=-\log x+\log c \end{aligned}

\begin{aligned} &\log \sqrt{\frac{v}{v+2}}+\log x=\log c \\ &\frac{x \sqrt{v}}{\sqrt{v+2}}=c \end{aligned}

Now putting value of v i.e , y/x

\begin{aligned} &\Rightarrow \frac{x \sqrt{\frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}

\begin{aligned} &\frac{\sqrt{x^{2} \times \frac{y}{x}}}{\sqrt{\frac{y}{x}+2}}=c \\ \end{aligned}

\begin{aligned} &\frac{\sqrt{x y}}{\sqrt{\frac{y+2 x}{x}}}=c \\ \end{aligned}

\begin{aligned} &\frac{x \sqrt{y}}{\sqrt{y+2 x}}=c \\ \end{aligned}

\begin{aligned} &x \sqrt{y}=c \sqrt{y+2 x} \end{aligned}

Squaring both sides

\begin{aligned} &x^{2} y=c^{2}(y+2 x) \quad \; \; \; \; \; \; \; \; \; \ldots (ii)\\ &\text { Now } x=1, y=1\\ &\text { Therefore }\left(1^{2}\right)(1)=c^{2}(1+2(1)) \end{aligned}

\begin{aligned} &\Rightarrow 1=c^{2}(3) \\ &\Rightarrow c^{2}=\frac{1}{3} \end{aligned}

Putting back in (ii)

\begin{aligned} &\Rightarrow x^{2} y=\frac{1}{3}(y+2 x) \\ &3 x^{2} y=y+2 x \\ &y+2 x=3 x^{2} y \end{aligned}